Derivative in cylindrical coordinates.

[yungman]

This is calculus question, but I don't think calculus really cover this topic in either multi-variables or even vector calculus classes. This is really more common problem in electrodynamics.

Let R be position vector that trace out a circle or radius a with constant velocity. In rectangular coordinates:

$$\vec R = \hat x acos \omega t +\hat y a sin\omega t$$

$$\vec v = \frac {d\vec R}{dt}= -\hat x a\omega sin \omega t +\hat y a\omega cos\omega t$$This is very straight forward. So I am going to try to do the derivative in cylindrical coordinates. For cylindrical coordinates;

$$\vec R =\hat r a$$

How do you take the derivative? There is only a constant. I know $$\vec v=\hat {\phi} a\omega$$

The only way I know how to get v in cylindrical coordinates is doing a translation of v from rectangular coordinates to cylindrical coordinate after performing the derivative in rectangular coordinates.

Anyone can tell me a direct method to take the derivative in cylindrical coordinates?

Also for constant angular velocity $\omega$, angular acceleration is supposed to be zero. But if you look at in rectangular coordinates:

$$\vec a = \frac {d \vec v}{dt} = -\hat x a \omega^2 cos \omega t \;-\; \hat y a\omega^2sin\omega t$$

Obviously it is not zero, does this mean you don't look at this as angular velocity or acceleration, but instead, look at it as in straight line components x,y?

 

[Vagn]

The absolute value of the acceleration would be constant for uniform circular motion as described, but wouldn't the unit vector be changing direction with respect to time. You'd need to find the time derivative of the unit vector in the r direction I think.

 

[yungman]

Thanks, that's what I suspect.

Anyone can help on the first part?

 

[nasu]

I am not sure what you call "the first part".
The idea is that in cylindrical coordinates the unit vectors are not constants, as suggested above.
The derivative of the position vector r is, in general
$$\frac{d \vec{r}}{dt}=\frac{d (r\hat{r})}{dt}= \hat{r} \frac{d r}{dt}+r \frac{d \hat{r}}{dt}$$
For circular motion the magnitude r is constant so only the second term remains.
The derivative of the unit vector r is
$$\frac{d (\hat{r})}{dt}=\omega \hat{\theta}$$

 

[yungman]

I am not sure what you call "the first part".
The idea is that in cylindrical coordinates the unit vectors are not constants, as suggested above.
The derivative of the position vector r is, in general
$$\frac{d \vec{r}}{dt}=\frac{d (r\hat{r})}{dt}= \hat{r} \frac{d r}{dt}+r \frac{d \hat{r}}{dt}$$
For circular motion the magnitude r is constant so only the second term remains.
The derivative of the unit vector r is
$$\frac{d (\hat{r})}{dt}=\omega \hat{\theta}$$

Thanks for the reply,

I just want to verify:

$$\hat r = \hat x\; cos\; \omega t + \hat y \;sin\;\omega t \; \Rightarrow \; \frac {d\hat r}{dt}= -\hat x\;\omega\; sin\; \omega t + \hat y\;\omega \;cos\;\omega t = \hat {\phi}\; \omega$$

Thanks

Alan

 

[nasu]

I think it's OK.

 

[yungman]

Thanks